\newproblem{lay:3_2_31}{
  % Problem identification
	\begin{large}
	  \hspace{\fill}\newline
    \textbf{Lay, 3.2.31}
	\end{large}
	\\
  \ifthenelse{\boolean{identifyAuthor}}{\textit{Carlos Oscar Sorzano, Aug. 31st, 2013} \\}{}

  % Problem statement
	Show that if $A$ is invertible, then $\det\{A^{-1}\}=\frac{1}{\det\{A\}}$
}{
   % Solution
	If $A$ is invertible, then
	\begin{center}
		$AA^{-1}=I$
	\end{center}
	Taking determinants on both sides
	\begin{center}
		$\det\{AA^{-1}\}=\det\{I\}$ \\
		$\det\{A\}\det\{A^{-1}\}=1$ \\
		$\det\{A^{-1}\}=\frac{1}{\det\{A\}}$
	\end{center}
}
\useproblem{lay:3_2_31}
\ifthenelse{\boolean{eachProblemInOnePage}}{\newpage}{}
